Anonymous user
Reduced row echelon form: Difference between revisions
→{{header|OCaml}}: (another implementation)
(Shorter D code) |
(→{{header|OCaml}}: (another implementation)) |
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Line 1,382:
=={{header|OCaml}}==
<lang
let tmp = m.(i) in
m.(i) <- m.(j);
Line 1,433:
print_newline()
) m</lang>
Another implementation:
<lang OCaml>let rref m =
let nr, nc = Array.length m, Array.length m.(0) in
let add r s k =
for i = 0 to nc-1 do m.(r).(i) <- m.(r).(i) +. m.(s).(i)*.k done in
for c = 0 to min (nc-1) (nr-1) do
for r = c+1 to nr-1 do
if r <> c && abs_float m.(c).(c) < abs_float m.(r).(c) then
let v = m.(r) in (m.(r) <- m.(c); m.(r) <- v)
done;
let t = m.(c).(c) in
if t <> 0.0 then
begin
for r = 0 to nr-1 do if r <> c then add r c (-.m.(r).(c)/.t) done;
for i = 0 to nc-1 do m.(c).(i) <- m.(c).(i)/.t done
end
done;;
let mat = [|
[| 1.0; 2.0; -.1.0; -.4.0;|];
[| 2.0; 3.0; -.1.0; -.11.0;|];
[|-.2.0; 0.0; -.3.0; 22.0;|]
|] in
let pr v = Array.iter (Printf.printf " %9.4f") v; print_newline() in
let show = Array.iter pr in
show mat;
print_newline();
rref mat;
show mat</lang>
=={{header|Octave}}==
| |||